need help with a trigonometric identity.

[kathi.s]

(cot theta / 1 - tan theta) + (tan theta / 1 - cot theta) = 1 + tan theta + cot theta


please please someone help me. I know it's something easy... I just can't figure out the steps. I'm positive I have to manipulate (tan theta / 1 - cot theta)


thank you!!

 

[soroban]

Hello, kathi.s!

\(\displaystyle \dfrac{\cot\theta}{1 - \tan\theta} + \dfrac{\tan\theta}{1 - \cot\theta} \;=\; 1 + \tan\theta + \cot\theta\)

We have: .\(\displaystyle \dfrac{\frac{1}{\tan\theta}}{1 - \tan\theta} + \dfrac{\tan\theta}{1 - \frac{1}{\tan\theta}} \)

Multiply by \(\displaystyle \frac{\tan\theta}{\tan\theta}:\;\;\dfrac{1}{\tan\theta(1-\tan\theta)} + \dfrac{\tan^2\theta}{\tan\theta - 1} \;\;=\;\;\dfrac{1}{\tan\theta(1-\tan\theta)} - \dfrac{\tan^2\theta}{1-\tan\theta} \)

Multiply the second fraction by \(\displaystyle \frac{\tan\theta}{\tan\theta}:\;\;\dfrac{1}{\tan\theta(1 - \tan\theta)} - \dfrac{\tan^3\theta}{\tan\theta(1-\tan\theta)} \)

. . \(\displaystyle =\;\;\dfrac{1-\tan^3\theta}{\tan\theta(1-\tan\theta)} \;\;=\;\;\dfrac{(1-\tan\theta)(1 + \tan\theta + \tan^2\theta)}{\tan\theta(1-\tan\theta)} \;\;=\) ..\(\displaystyle \dfrac{1 + \tan\theta + \tan^2\theta}{\tan\theta} \)

. . \(\displaystyle =\;\;\dfrac{1}{\tan\theta} + \dfrac{\tan\theta}{\tan\theta} + \dfrac{\tan^2\theta}{\tan\theta} \;\;=\;\; \cot\theta + 1 + \tan\theta \)

 

[kathi.s]

Thank you so much!!!
What i need clarity on is this: on the second fraction, how you go from [(tan x) /( 1 -( 1 / tan x))] to [(tan^2 x) / (tan x - 1)]

thank you once again

 

[kathi.s]

NEVERMIND!!! i got it!
thank you
!